# Is the $H$-space structure on $S^7$ associative up to homotopy?

Endow $S^7$ with a structure of an $H$-space induced from multiplication in octonions $\mathbb{O}=\mathbb{R}^8$. It is not associative as octonion multiplication is not associative.

Is it associative up to homotopy, i.e. are maps $m(m(-,-),-):S^7\times S^7\times S^7\to S^7$ and $m(-,m(-,-)):S^7\times S^7\times S^7\to S^7$ homotopic?

It is not. See Theorem 1.4 of this paper by I.M. James (Trans. AMS 84 (1957), 545-558).

In particular, there exists no homotopy associative multiplication on $S^n$ unless $n=1$ or $n=3$.

• Or n=0, of course. – Theo Johnson-Freyd Dec 4 '16 at 3:25
• Multiplications on $S^7$ are in 1-1 correspondence with the elements of $\pi_{14}S^7=\mathbb{Z}_{8}\oplus\mathbb{Z}_{3}\oplus\mathbb{Z}_{5}$. Given a product $\mu:S^7\times S^7\rightarrow S^7$, the obstruction it being homotopy associative lies in $\pi_{21}S^7=\mathbb{Z}_{8}\oplus\mathbb{Z}_{4}\oplus\mathbb{Z}_{3}$. As Jon says, there are no homotopy associative multiplications on $S^7$. In fact $S^7$ has no 2-local or 3-local associative products. One must invert both the primes 2 and 3 before $S^7$ admits a homotopy associative multiplication (and then any multiplication will be HA). – Tyrone Dec 4 '16 at 15:59
• @Tyrone A question arises which I wanted to formulate rigorously but failed - is it still "homotopy something"? That is, is the existing multiplication completely generic in some sense? – მამუკა ჯიბლაძე Dec 5 '16 at 5:47
• @მამუკაჯიბლაძე The octonions form an alternative algebra, i.e. $x(xy) = (xx)y$ and $(yx)x = y(xx)$ (and $S^7$ is the unit sphere of the algebra of octonions as far as I can tell). – Najib Idrissi Dec 5 '16 at 8:42
• @ConnorMalin, correct. You need a multiplication $\mu$ to begin with, then you get a new one by defining $\mu'=\mu+\alpha\circ q$, (as elements of the algebraic loop $[X\times X,X]$) where $\alpha:X\wedge X\rightarrow X$ and $q$ is the quotient to the smash. Since the only spheres that admit (integral) multiplications are $S^1,S^3,S^7$, among spheres the statement is quite special to these chaps. – Tyrone Jul 25 '19 at 18:03

There is a Proof due to Stasheff in "H-space from homotopy point of view" (Theorem 6.7). The argument is fairly simple to describe. $\def\OP{{\mathbb O\mathbf P}}$

If $S^7$ admits a homotopy associative multiplication, then one should be able to construct $\OP^3$. It would follow that $\tilde{H}^*(\OP^3;\mathbb{Z}/3)$ is generated by $u_8$ in degree $8$ with the relation that $u_8^4 =0$. It follows that, $$P^4(u_8) = u_8^3.$$ However, $P^4 = P^1P^3$, which means the $u_8^3 = P^1x$, where $x \in \tilde{H}^{20}(\OP^3;\mathbb{Z}/3) = 0$. Thus $u_8^3 = 0$ which contradicts the existence of $\OP^3$ and consequently the existence of homotopy associative multiplication on $S^7$.

More interestingly this proof suggests that the obstruction to homotopy associative multiplication is $3$-local. This leads to the following question.

Question: Is $S^7_{(2)}$ homotopy associative? Does the reference to James work in the answer due to Jon Beardsley, addresses this question? (Tyrone's comment made me think of this question.)

• Isn't the associativity itself required to construct the classifying space which turns out to have the same cohomology as $\mathbb{OP}^{\infty}$? (assuming I got the argument right) – SashaP Dec 5 '16 at 18:37
• What you need is $A_{\infty}$-structure (slightly weaker than strict associativity) to construct classifying space/bar complex. Stasheff showed that you can constructed an $n$-truncated bar-complex for an $H$-space which is $A_n$. In particular homotopy associativity is equivalent to $A_3$-structure. If $S^7$ admits $A_3$-structure then you construct the $3$-truncated bar-complex aka $\mathbb{OP}^{3}$, and that is all is required for the proof above. – Prasit Dec 5 '16 at 19:20